# Number of divisors of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form $4n+1,n\in N$?

In this question, the OP asks to find the number of divisors of $$2^2\cdot 3^3\cdot 5^3\cdot 7^5$$ which are of the form $$4n+1,n\in N$$. The top answer points out that the required divisor is of the form $$3^a\cdot 5^b\cdot 7^c$$ with $$0\leq a\leq 3,0\leq b\leq 3,0\leq c\leq 5$$ and $$a+c$$ being even. The answer therefore is, apparently, $$(4 \cdot 4 \cdot 6)/2=48$$.

But this is wrong according to my book: the correct answer is $$47$$. Obviously, one case has been overcounted, but which? As far as I know, the person who wrote the top answer employed a fairly standard approach and should have arrived at the correct answer.

• I guess the book uses $0 \notin \mathbb{N}$, so it doesn't count $1 = 4\cdot 0 + 1$. – Daniel Fischer Nov 20 '20 at 16:18
• Well, at a guess, the official solution excludes $(a,b,c)=(0,0,0)$, i.e. it excludes the divisor $1$. – lulu Nov 20 '20 at 16:18
• @Daniel Fischer Ah, that makes sense. Thanks. – Ray Bradbury Nov 20 '20 at 16:21
• Does this answer your question? Number of divisors of the form $(4n+1)$ – Dave Nov 22 '20 at 7:41

As pointed out in the comments by Daniel Fischer and lulu, my book considers $$0 \notin N$$, so discounts the case where $$a=b=c=0$$, i.e., $$4(0)+1=1$$.